Question

A small rock is thrown straight up from the edge of the roof of an 8.00 m tall building with an initial speed vo . The speed of the rock just before it strikes the ground is 24.0 m/s. What is the initial speed, vo , of the rock?

Answer 1

Answer:

20.47 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{24^2-0^2}{2\times 9.81}\\\Rightarrow s=29.3577\ m$

Total height of the fall is 29.3577 m

Height the ball reached above the building is $29.3577-8=21.3577\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21.3577=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{21.3577\times 2}{9.81}}\\\Rightarrow t=2.0866\ s$

Time taken to reach the point from where the ball was thrown is 2.0866 s

This will also be the time it takes the ball to reach the maximum height

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\dfrac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\dfrac{21.3577-\frac{1}{2}\times -9.81\times 2.0866^2}{2.0866}\\\Rightarrow u=20.47\ m/s$

The initial velocity with which the rock was thrown was 20.47 m/s